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Dec 21, 2021 at 12:43 vote accept dohmatob
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Dec 6, 2018 at 14:29 answer added Aryeh Kontorovich timeline score: 1
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Aug 8, 2018 at 7:43 comment added dohmatob @MateuszKwaśnicki OK, Thanks. The issue you raise is more fundamental / crucial than I initially thought. Let me try to fix it. (It may turn out that I'm not asking the right question at all...)
Aug 8, 2018 at 7:42 history edited dohmatob CC BY-SA 4.0
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Aug 8, 2018 at 7:34 history edited dohmatob CC BY-SA 4.0
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Aug 8, 2018 at 7:32 comment added Mateusz Kwaśnicki Does this really change anything? You can now replace $h$ by $\tfrac{1}{2} + a (h - \tfrac{1}{2})$ to get the same $c_h$ with a smaller Lipschitz constant.
Aug 8, 2018 at 7:23 comment added dohmatob @MateuszKwaśnicki Thanks! You're completely right. I screwed things when I tried to map my problem from $\{0,1\}$ to $\{-1,1\}$. Fixed. Let me know if this works for you.
Aug 8, 2018 at 7:21 history edited dohmatob CC BY-SA 4.0
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Aug 8, 2018 at 7:08 history edited dohmatob CC BY-SA 4.0
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Aug 8, 2018 at 7:08 comment added Mateusz Kwaśnicki I do not think the answer depends on $L$ at all: you can always replace $h$ by $a h$ with $a \in (0, 1)$ to get a Lipschitz function with a smaller Lipschitz constant $a L$. Am I missing something?
Aug 8, 2018 at 7:07 history edited dohmatob CC BY-SA 4.0
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Aug 8, 2018 at 7:00 history edited dohmatob CC BY-SA 4.0
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Aug 6, 2018 at 20:09 answer added usul timeline score: 1
Aug 6, 2018 at 18:11 history edited dohmatob CC BY-SA 4.0
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Aug 6, 2018 at 17:25 history edited dohmatob CC BY-SA 4.0
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Aug 6, 2018 at 16:22 history edited dohmatob CC BY-SA 4.0
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Aug 6, 2018 at 16:16 history edited dohmatob CC BY-SA 4.0
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Aug 6, 2018 at 15:38 history asked dohmatob CC BY-SA 4.0